What is expand ((2x+x^3)^3)/(12) ?

The solution to expand ((2x+x^3)^3)/(12) is (x^9}{12}+\frac{x^7)/2+x^5+(2x^3)/3

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expand ((2x+x^3)^3)/(12)

expand

\frac{( 2 x + x ^{3} ) ^{3}}{12}

\frac{x ^{9}}{12} + \frac{x ^{7}}{2} + x^{5} + \frac{2 x ^{3}}{3}

Solution steps

\frac{( 2 x + x ^{3} ) ^{3}}{12}

(2 x + x^{3})^{3} = 8 x^{3} + 12 x^{5} + 6 x^{7} + x^{9}

= \frac{8 x ^{3} + 12 x ^{5} + 6 x ^{7} + x ^{9}}{12}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{8 x ^{3} + 12 x ^{5} + 6 x ^{7} + x ^{9}}{12} = \frac{8 x ^{3}}{12} + \frac{12 x ^{5}}{12} + \frac{6 x ^{7}}{12} + \frac{x ^{9}}{12}

= \frac{8 x ^{3}}{12} + \frac{12 x ^{5}}{12} + \frac{6 x ^{7}}{12} + \frac{x ^{9}}{12}

Cancel

\frac{8 x ^{3}}{12} : \frac{2 x ^{3}}{3}

= \frac{2 x ^{3}}{3} + \frac{12 x ^{5}}{12} + \frac{6 x ^{7}}{12} + \frac{x ^{9}}{12}

Divide the numbers:

\frac{12}{12} = 1

= \frac{2 x ^{3}}{3} + x^{5} + \frac{6 x ^{7}}{12} + \frac{x ^{9}}{12}

Cancel

\frac{6 x ^{7}}{12} : \frac{x ^{7}}{2}

= \frac{2 x ^{3}}{3} + x^{5} + \frac{x ^{7}}{2} + \frac{x ^{9}}{12}

Rewrite in standard form

= \frac{x ^{9}}{12} + \frac{x ^{7}}{2} + x^{5} + \frac{2 x ^{3}}{3}

(y^{2} + 3 x) d x + (e^{x}) d y

(1, i) (2 - 1, 3 + 2 i)

\frac{( x + 3 ) ( 3 x + 1 + 1 )}{3}

e^{x} (x^{3} - 3 x)

\frac{1}{x ( a - b ln ( x ) )}

What is expand ((2x+x^3)^3)/(12) ?
The solution to expand ((2x+x^3)^3)/(12) is (x^9}{12}+\frac{x^7)/2+x^5+(2x^3)/3